E=energy=5.09x10^5J = 509KJ
<span>M=mass=2250g=2.25Kg </span>
<span>C=specific heat capacity of water= 4.18KJ/Kg </span>
<span>ΔT= change in temp= ? </span>
<span>E=mcΔT </span>
<span>509=(2.25)x(4.18)xΔT </span>
<span>509=9.405ΔT </span>
<span>ΔT=509/9.405=54.1degrees </span>
<span>Initial temp = 100-54 = 46 degrees </span>
<span>Hope this helps :)</span>
Answer:
It's held together by the nuclear force.
Explanation:
There are <em>more</em> elemental forces than just the electromagnetic one. In this case, it is the nuclear force (called also strong force) the one that holds the nucleus together because it is stronger than the electromagnetic force over such short distances as the one inside the atomic nucleus.
Answer:
a

b

Explanation:
From the question we are told that
The mass of the rock is 
The length of the small object from the rock is 
The length of the small object from the branch 
An image representing this lever set-up is shown on the first uploaded image
Here the small object acts as a fulcrum
The force exerted by the weight of the rock is mathematically evaluated as

substituting values


So at equilibrium the sum of the moment about the fulcrum is mathematically represented as

Here
is very small so
and 
Hence

=> 
substituting values


The mechanical advantage is mathematically evaluated as

substituting values

